Problem: Is ${691962}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {691962}= &&{6}\cdot100000+ \\&&{9}\cdot10000+ \\&&{1}\cdot1000+ \\&&{9}\cdot100+ \\&&{6}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {691962}= &&{6}(99999+1)+ \\&&{9}(9999+1)+ \\&&{1}(999+1)+ \\&&{9}(99+1)+ \\&&{6}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {691962}= &&\gray{6\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {6}+{9}+{1}+{9}+{6}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${691962}$ is divisible by $9$ if ${ 6}+{9}+{1}+{9}+{6}+{2}$ is divisible by $9$ Add the digits of ${691962}$ $ {6}+{9}+{1}+{9}+{6}+{2} = {33} $ If ${33}$ is divisible by $9$ , then ${691962}$ must also be divisible by $9$ ${33}$ is not divisible by $9$, therefore ${691962}$ must not be divisible by $9$.